The stress-energy tensor of a perfect fluid is given by

$$T^{\mu\nu}=\left(\rho+pc^{-2}\right)u^\mu u^\nu+pg^{\mu\nu}$$

The divergence of the stress-energy tensor is zero: $\nabla_\mu T^{\mu\nu}=0$. Hence

$$\nabla_\mu\left(\rho+pc^{-2}\right)u^\mu u^\nu+\nabla_\mu pg^{\mu\nu}=0$$

Expanding the first term, and using the product rule on the second term, yields

$$\nabla_\mu\rho u^\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\left(\nabla_\mu p\right)g^{\mu\nu}+p\nabla_\mu g^{\mu\nu}=0$$

Using the product rule once again on the first term yields

$$\left(\nabla_\mu\rho u^\mu\right) u^\nu+\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\left(\nabla_\mu p\right)g^{\mu\nu}+p\nabla_\mu g^{\mu\nu}=0$$

By the continuity equation, $\nabla_\mu\rho u^\mu=0$. Hence

$$\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\left(\nabla_\mu p\right)g^{\mu\nu}+p\nabla_\mu g^{\mu\nu}=0$$

The divergence of the metric tensor is zero: $\nabla_\mu g^{\mu\nu}=0$. Hence

$$\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\left(\nabla_\mu p\right)g^{\mu\nu}=0$$

Finally, using tensor contraction on the last term yields

$$\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\nabla^{\nu}p=0$$

We now turn to the Cauchy momentum equation in the Euler equations:

$$0=\rho\left(\frac{\partial}{\partial t}+\vec{u}\cdot\vec{\nabla}\right)\vec{u}+\vec{\nabla}p= \rho\left(c\nabla_0+\vec{u}\cdot\vec{\nabla}\right)\vec{u}+\vec{\nabla}p$$

Using the non-relativistic approximation $\gamma\approx1$ we obtain:

$$0\approx \rho\left(\gamma c\nabla_0+\gamma\vec{u}\cdot\vec{\nabla}\right)\gamma\vec{u}+\vec{\nabla}p= \rho u^{\mu}\nabla_{\mu}u^i+\nabla^ip$$

Compare this with the result obtained from the stress-energy tensor:

$$0=\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\nabla^{\nu}p$$

Why is there an extra term ($\nabla_\mu pc^{-2}u^\mu u^\nu$)? Does it vanish in the non-relativistic limit, simply because of the $c^{-2}$ factor?

  • $\begingroup$ To take the relativistic limit of the energy-momentum conservation equation, make the following approximation: $u^i$ small, $u^0\approx 1$, $\partial_\lambda u^\mu= \mathcal O (u^i)\; \forall \mu,\lambda;\; p=\mathcal O(u^i)$, and keep only first order terms in $u^i$. This leads to the right equation, at least for me. $\endgroup$
    – Danu
    May 25, 2015 at 8:53

1 Answer 1


Note that $\nabla_\mu (\rho u^\mu)=0$ is not correct (and not the continuity equation). Keep in mind that (based on your definition of the stress tensor) $\rho$ is the energy density, and the conserved energy current is $T_{0\mu}$.

The relativistic Euler (momentum conservation) equation is $$ D u_\mu = -\frac{1}{\rho+P}\nabla_\mu^\perp P $$ where $D=u^\mu\nabla_\mu$ and $\nabla_\mu^\perp=(g_{\mu\nu}-u_\mu u_\nu)\nabla^\nu$, which shows that the relativistic inertia is the enthalpy density, $w=\rho+P$. In the non-relativistic limit, $u_0\simeq 1$, $D$ is the comoving derivative, $w$ is the mass density, and $\nabla_i^\perp\simeq\nabla_i$. This leads to the standard Euler equation.

  • 1
    $\begingroup$ I'm pretty sure $\nabla_\mu(\rho u^\mu) = 0$ is the correct continuity equation. It holds in arbitrary spacetimes for any perfect fluid with conserved rest mass density $\rho$. $\endgroup$
    – user10851
    May 26, 2015 at 4:28
  • 1
    $\begingroup$ No. If there is a conserved particle number, and if $\rho$ is the density of particles, and if $u_\mu$ is defined to be the velocity of the particles (the so-called Eckart frame), then this equation would be correct. But, as is clear from the expression for $T_{\mu\nu}$, $\rho$ is the energy density. $\endgroup$
    – Thomas
    May 26, 2015 at 5:15
  • $\begingroup$ It has to be energy density, otherwise the formula for the stress tensor is wrong. $\endgroup$
    – Thomas
    May 26, 2015 at 20:06
  • $\begingroup$ Terminology seems to be a problem here. Refer to section 3.2 ("Stress energy tensor") in mathreview.uwaterloo.ca/archive/voli/2/olsthoorn.pdf. It states that both $T^{\mu\nu}=(\rho+p)u^{\mu}u^{\nu}+pg^{\mu\nu}$ (assuming $\epsilon=0$) and $\nabla_\mu(\rho u^{\mu})=0$, so both equations are correct. $\endgroup$
    – user76284
    May 26, 2015 at 22:44
  • $\begingroup$ I think this is not just terminology. $T_{\mu\nu}=(\rho+P)u_\mu u_\nu+Pg_{\mu\nu}$ and $j_\mu=\rho u_\mu$ cannot both be correct. You can consider $\epsilon=0$ (in your notation) but i) there is no such fluid known to man (finite pressure but zero internal energy), ii) the original question is pointless, because we are in the non-relativistic limit from the start. $\endgroup$
    – Thomas
    May 27, 2015 at 2:48

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